Answer
(a) The series converges.
(b) The series diverges.
Work Step by Step
(a) Write $\mathop \sum \limits_{k = 2}^\infty \left[ {\dfrac{1}{{k{{\left( {\ln k} \right)}^2}}} - \dfrac{1}{{{k^2}}}} \right] = \mathop \sum \limits_{k = 2}^\infty \dfrac{1}{{k{{\left( {\ln k} \right)}^2}}} - \mathop \sum \limits_{k = 2}^\infty \dfrac{1}{{{k^2}}}$.
(i) Consider the series $\mathop \sum \limits_{k = 2}^\infty \dfrac{1}{{k{{\left( {\ln k} \right)}^2}}}$.
The series has positive terms.
Write ${a_k} = f\left( k \right)$, where $f\left( x \right) = \dfrac{1}{{x{{\left( {\ln x} \right)}^2}}}$.
The function $f$ is continuous and decreasing for $x \ge 2$. Thus, the integral test is applicable.
Evaluate $\mathop \smallint \limits_2^\infty \dfrac{1}{{x{{\left( {\ln x} \right)}^2}}}{\rm{d}}x$.
Let $t = \ln x$. So, $dt = \dfrac{1}{x}dx$. The integral becomes
$\mathop \smallint \limits_2^\infty \dfrac{1}{{x{{\left( {\ln x} \right)}^2}}}{\rm{d}}x = \mathop {\lim }\limits_{p \to \infty } \mathop \smallint \limits_{\ln 2}^{\ln p} \dfrac{1}{{{t^2}}}{\rm{d}}t = - \mathop {\lim }\limits_{p \to \infty } \left( {\dfrac{1}{t}} \right)|_{\ln 2}^{\ln p} = - \mathop {\lim }\limits_{p \to \infty } \left( {\dfrac{1}{{\ln p}} - \dfrac{1}{{\ln 2}}} \right) = \dfrac{1}{{\ln 2}}$
Since $\mathop \smallint \limits_2^\infty \dfrac{1}{{x{{\left( {\ln x} \right)}^2}}}{\rm{d}}x$ converges, by Theorem 9.4.4, the series $\mathop \sum \limits_{k = 2}^\infty \dfrac{1}{{k{{\left( {\ln k} \right)}^2}}}$ also converges.
(ii) Consider the series $\mathop \sum \limits_{k = 2}^\infty \dfrac{1}{{{k^2}}}$.
The series converges because it is a $p$-series with $p = 2 > 1$.
Since both the series $\mathop \sum \limits_{k = 2}^\infty \dfrac{1}{{k{{\left( {\ln k} \right)}^2}}}$ and $\mathop \sum \limits_{k = 2}^\infty \dfrac{1}{{{k^2}}}$ converge, by Theorem 9.4.3 (a), the series $\mathop \sum \limits_{k = 2}^\infty \left[ {\dfrac{1}{{k{{\left( {\ln k} \right)}^2}}} - \dfrac{1}{{{k^2}}}} \right]$ converges.
(b) Write $\mathop \sum \limits_{k = 2}^\infty \left[ {k{{\rm{e}}^{ - {k^2}}} + \dfrac{1}{{k\ln k}}} \right] = \mathop \sum \limits_{k = 2}^\infty k{{\rm{e}}^{ - {k^2}}} + \mathop \sum \limits_{k = 2}^\infty \dfrac{1}{{k\ln k}}$.
(i) Consider the series $\mathop \sum \limits_{k = 2}^\infty \dfrac{k}{{{{\rm{e}}^{{k^2}}}}}$. The series has positive terms.
Write ${a_k} = f\left( k \right)$, where $f\left( x \right) = \dfrac{x}{{{{\rm{e}}^{{x^2}}}}}$.
The function $f$ is continuous and decreasing for $x \ge 2$. Thus, the integral test is applicable.
Evaluate $\mathop \smallint \limits_2^\infty \dfrac{x}{{{{\rm{e}}^{{x^2}}}}}{\rm{d}}x$.
Let $t = {x^2}$. So, $dt = 2xdx$. The integral becomes
$\mathop \smallint \limits_2^\infty \dfrac{x}{{{{\rm{e}}^{{x^2}}}}}{\rm{d}}x = \dfrac{1}{2}\mathop \smallint \limits_4^\infty \dfrac{1}{{{{\rm{e}}^t}}}{\rm{d}}t = - \dfrac{1}{2}\mathop {\lim }\limits_{p \to \infty } \left( {{{\rm{e}}^{ - t}}} \right)|_4^{{p^2}} = - \dfrac{1}{2}\mathop {\lim }\limits_{p \to \infty } \left( {{{\rm{e}}^{ - {p^2}}} - {{\rm{e}}^{ - 4}}} \right) = \dfrac{1}{{2{{\rm{e}}^4}}}$
Since $\mathop \smallint \limits_2^\infty \dfrac{x}{{{{\rm{e}}^{{x^2}}}}}{\rm{d}}x$ converges, by Theorem 9.4.4, the series $\mathop \sum \limits_{k = 2}^\infty \dfrac{k}{{{{\rm{e}}^{{k^2}}}}}$ also converges.
(ii) Consider the series $\mathop \sum \limits_{k = 2}^\infty \dfrac{1}{{k\ln k}}$. The series has positive terms.
Write ${a_k} = f\left( k \right)$, where $f\left( x \right) = \dfrac{1}{{x\ln x}}$.
The function $f$ is continuous and decreasing for $x \ge 2$. Thus, the integral test is applicable.
Evaluate $\mathop \smallint \limits_2^\infty \dfrac{1}{{x\ln x}}{\rm{d}}x$.
Let $t = \ln x$. So, $dt = \dfrac{1}{x}dx$. The integral becomes
$\mathop \smallint \limits_2^\infty \dfrac{1}{{x\ln x}}{\rm{d}}x = \mathop {\lim }\limits_{p \to \infty } \mathop \smallint \limits_{\ln 2}^{\ln p} \dfrac{1}{t}{\rm{d}}t = \mathop {\lim }\limits_{p \to \infty } \left( {\ln t} \right)|_{\ln 2}^{\ln p} = \mathop {\lim }\limits_{p \to \infty } \left( {\ln \left( {\ln p} \right) - \ln \left( {\ln 2} \right)} \right) = \infty $
Since $\mathop \smallint \limits_2^\infty \dfrac{1}{{x\ln x}}{\rm{d}}x$ diverges, by Theorem 9.4.4, the series $\mathop \sum \limits_{k = 2}^\infty \dfrac{1}{{k\ln k}}$ also diverges.
(iii) We have $\mathop \sum \limits_{k = 2}^\infty k{{\rm{e}}^{ - {k^2}}}$ converges and $\mathop \sum \limits_{k = 2}^\infty \dfrac{1}{{k\ln k}}$ diverges. According to Exercise 27, the series $\mathop \sum \limits_{k = 2}^\infty \left[ {k{{\rm{e}}^{ - {k^2}}} + \dfrac{1}{{k\ln k}}} \right]$ diverges.
